MAC 2312
INFINITE SERIES PROBLEMS MIXED TOGETHER
Determine if the given series converges or diverges.
∞
1)
∑ 6n
n=1
n
+2
3
Since an is a rational expression, rule out the Ratio Test or Root Test and consider the
∞
n
n
1
1
< 3 = 2 and ∑ 2 converges (p series with p>1), we have
Comparison Test. Noting that an = 3
6n + 2 n
n
n=1 n
∞
n
is convergent.
n=1 6n + 2
that by the Comparison Test ∑
3
∞
2)
5n+3
∑ 82n
n=1
∞
⎛ 5n ⎞
5
5
5n+3
5n+3
3
< 1⇒ by the
is a geometric series with common ratio r =
. r =
=
5
⇒
∑
2n
n⎟
2n
⎜
64
64
8
⎝ 64 ⎠
n=1 8
Geometric Series Test, the series converges.
an =
∞
3)
∑
(n + 1)
2
⋅ 4n
n!
n=1
The presence of a factorial term in an suggests the use of the Ratio Test.
( n + 2) ⋅ 4
(n + 1)!
=
(n + 1) ⋅ 4
2
an+1
an
n+1
2
( n + 2) ⋅ 4
=
(n + 1)!
2
n
n+1
⋅
(
n!
)
2
n + 1 ⋅ 4n
=
(
4 n+2
(
)(
)
2
=
) (n
n +1 n +1
2
n!
4n 2 + 16n + 16
4
ρ = lim 3
= lim = 0. ρ < 1 so by the Ratio Test,
2
n→∞ n + 3n + 3n + 1
n→∞ n
4)
∞
∑
n=1
∞
∑
n=1
(
4 n 2 + 4n + 4
3
)
)
+ 3n + 3n + 1
2
(n + 1)
2
n!
⋅ 4n
⇒
is convergent.
5
n−3
5
n−3
Test,
>
∞
∑
n=1
1
n
5
=
n−3
1
and
n1/2
∞
∑
n=1
diverges.
1
n
diverges because it is a p-series with p < 1. So, by the Comparison
∞
5)
9n 2
∑ 7n3 − 1
n=1
9n 2
9 1
>
> and
3
7n − 1 7n n
6)
∞
nn ⋅ n!
∑ 2n !
n=1
( )
∞
1
∑n
∞
diverges. So, by the Comparison Test,
n=1
9n 2
diverges.
3
−1
∑ 7n
n=1
Use the Ratio Test.
(n + 1) ⋅ (n + 1)!
⎡ 2 ( n + 1) ⎤ !
a
(n + 1) ⋅ (n + 1)! ⋅ ( 2n)! = (n + 1) ⋅ (n + 1)! ⋅
( 2n)!
⎣
⎦
=
=
=
a
n!
n ⋅ n!
( 2n + 2)! n ⋅ n! n
( 2n + 2)( 2n + 1)( 2n)!
( 2n)!
(n + 1) (n + 1) ⋅ (n + 1)! ⋅
⎛ n + 1⎞ ( n + 1) ( n + 1)
⎛
1
1 ⎞ ⎛ n + 2n + 1 ⎞
=⎜
= ⎜ 1+ ⎟ ⎜
⎟
n!
n
( 2n + 2)( 2n + 1) ⎝ n ⎠ ( 2n + 2)( 2n + 1) ⎝ n ⎠ ⎝ 4n + 6n + 2 ⎟⎠
n+1
n+1
n+1
n+1
n
n
n
n
n
n
n
n
2
2
n
n
⎡⎛
⎛
1 ⎞ ⎛ n 2 + 2n + 1 ⎞ ⎤
n 2 + 2n + 1
1⎞
1
e
⎥ = lim 2
ρ = lim ⎢⎜ 1+ ⎟ ⎜ 2
⋅ lim ⎜ 1+ ⎟ = e =
⎟
n→∞ ⎝
n ⎠ ⎝ 4n + 6n + 2 ⎠ ⎥ n→∞ 4n + 6n + 2 n→∞ ⎝ n ⎠
4
4
⎢⎣
⎦
e
< 1⇒ ρ < 1⇒
4
x
⎛
1⎞
from use of the definition lim ⎜ 1+ ⎟ = e and some annoying algebra, we have that the series
x→∞ ⎝
x⎠
converges by the Ratio Test.
∞
7)
n3
∑ n3 + 3n2 + 3n + 1
n=1
n3
= 1 lim an ≠ 0 ⇒
n→∞
n→∞ n 3 + 3n 2 + 3n + 1
n→∞
(aka The Divergence Test)
lim an = lim
⎛ π⎞
8) ∑ ⎜ − ⎟
⎝ e⎠
∞
∞
n3
∑ n3 + 3n2 + 3n + 1 diverges by the Nth Term Test
n=1
n+1
n=0
This is a geometric series with common ratio r = −
⎛ π⎞
Test, ∑ ⎜ − ⎟
e⎠
n=0 ⎝
∞
n+1
diverges
π
. Since r > 1 , we have that by the Geometric Series
e
∞
5n 2 + 2
∑ n4 + 4n
n=1
9)
5n 2 + 2 5n 2 + n 2 6n 2 6
<
= 4 = 2 and
n 4 + 4n
n4
n
n
∞
6
∑n
n=1
2
converges (p-series test with p=2). So, by the Comparison
∞
5n 2 + 2
Test, ∑ 4
converges
n=1 n + 4n
10)
∞
∑
4 + cos x
n3
n=1
4 4 + cos x
5 ∞ 5
≤ 3 . ∑ 3 is a convergent p-series. So,
For any x, 0 ≤ cos x ≤ 1⇒ 3 ≤
n
n3
n n=1 n
converges and by the Comparison Test.
∞
11)
∑ ne
∞
∑
4 + cos x
n=1
n3
also
−n
n=1
Positive term series whose nth terms contain exponential terms or logarithms are possibly good
candidates for the Integral Test. Let f x = xe − x . After checking that f satisfies the hypotheses of the
()
∞
Integral Test , compute
∫ xe
−x
.
1
∫ xe
−x
dx = uv − ∫ v du = −xe
u=x dv =e − xdx
du=dx v =−e − x
(
lim −xe − x − e − x
c→∞
− lim
c→∞
)
c
1
(
−x
− ∫ −e dx = − xe
−x
) (
)
−x
+ ∫e
−x
= −xe
−x
−e
−x
∞
+ C ⇒ ∫ xe − x =
1
(
c +1
=
+1
c→∞ e c L'Hopital's Rule
)
= lim −ce −c − e −c − 0 − 1 = − lim ⎡⎣e −c c + 1 ⎤⎦ + 1= − lim
c→∞
c→∞
1
+ 1= 0 + 1= 1< ∞
ec
∞
Since the integral converges, then so does the series ∑ ne −n .
n=1
12)
∞
∑
n=2
1
( )
n lnn
3
( )
As with #12, try the Integral Test. First, confirm that f x =
Integral Test.
1
( )
x ln x
3
satisfies the hypotheses of the
u −2
1
1
∫ x ln x 3 dx = ∫ u dudx = −2 + C = − 2u 2 + C = − 2 ln x
u=ln x du=
1
−3
( )
( )
x
⎛
1
lim ⎜ −
c→∞
⎜⎝ 2 ln x
( )
c
⎞
1
1
⎟ = − lim
2
2 c→∞ lnc
⎟⎠
2
( )
So, by the Integral Test,
∞
∑
n=1
∞
13)
⎛
1
−⎜−
⎜⎝ 2 ln2
( )
2
1
( )
n lnn
3
∞
2
+C ⇒ ∫
2
⎞
1
⎟ = 0+
2
⎟⎠
2 ln2
( )
2
=
1
( )
x ln x
( )
dx =
∞
1
2 ln2
3
2
<∞⇒∫
2
1
( )
x ln x
3
dx
is a convergent series.
n2
2
−1
∑ 9n
n=1
n2
1
=
2
n→∞9 9n − 1
9
lim an = lim
n→∞
∞
n2
diverges by the Nth Term Test.
2
n=1 9n − 1
lim an ≠ 0 ⇒ ∑
n→∞
tan−1 n
∑ 1+ n2
n=1
∞
14)
SOLUTION ONE: Use the Integral Test
( )
f x =
−1
tan x
⇒
1+ x 2
( tan x )
lim
−1
c→∞
2
c
−1
tan x
∫ 1+ x 2 dx =
1
u=tan−1 x du=
=
1
1+x 2
dx
u
2
( tan x )
+C =
−1
2
2
∞
tan−1 x
dx =
2
1+
x
1
+C ⇒ ∫
2
2
∞
2
2
1
1
1 ⎡⎛ π ⎞ ⎛ π ⎞ ⎤ 3π 2
tan−1 x
−1
−1
= lim tan x − tan 1 = ⎢⎜ ⎟ − ⎜ ⎟ ⎥ =
<∞⇒∫
dx
2 c→∞
2
2 ⎢⎝ 2 ⎠ ⎝ 4 ⎠ ⎥ 32
1+ x 2
1
⎣
⎦
(
2
∫ u du
2
)
(
)
tan−1 n
converges ⇒ ∑
converges.
2
n=1 1+ n
∞
∞
SOLUTION TWO: Use the Comparison Test with
1
∑n
n=1
2
. This can be done because the range of
⎛ π π⎞
tan−1 x is ⎜ − , ⎟
⎝ 2 2⎠
Each series below is a geometric series. Identify the common ratio of the series and then decide if the
series is convergent or divergent.
∞
15)
2 ⋅ 5n−1
∑ 11n−2
n=1
⎛ 5⎞
2 ⋅ 5n−1 2 ⋅ 5 ⋅ 5n−2
=
= 10 ⎜ ⎟
n−2
n−2
⎝ 11⎠
11
11
n−2
⇒r =
2 ⋅ 50
1st term 111−2
The sum S =
=
=
5
1− r
1−
11
5
11
r < 1⇒ The series converges.
2
11−1 = 11⋅ 2 ⋅ 11 = 121
6
6
3
11
∞
23n
16) ∑ 2n+1
n=1 3
n
23n
8n
1 ⎛ 8⎞
8
=
= ⎜ ⎟ ⇒r =
2n+1
n
9
3
9 ⋅3 3 ⎝ 9⎠
r < 1⇒ The series converges.
8
8
1st term
8 9 8
The sum S =
= 27 = 27 =
⋅ =
8
1
1− r
27 1 3
1−
9
9
∞
17)
∑ ( −0.9999)
n
n= 2
r = −0.9999
r < 1⇒ The series converges.
(
)
2
−0.9999
1st term
The sum S =
=
1− r
1− −0.9999
∞
18)
∑ (sec 4)
n
r = sec 4
(
)
Simplifying is not required here.
r ≥ 1⇒ The series diverges. There is no sum.
n=1
∑ (log
∞
19)
n=1
2
2
)
The sum S =
∞
20)
1
2
n−1
∑ (ln3)
n−1
r = log2 2 = log2 2 =
1st term
=
1− r
1
1
1−
2
1
2
r < 1⇒ The series converges.
=2
r = ln3 < lne = 1⇒ The series diverges. There is no sum.
n=1
Decide whether the given statement is true or false. Explain your reasoning.
∞
21) If
∑a
k
k =1
TRUE
∞
and
∑b
k
k =1
∞
are convergent, then so is
∑ (a
k
k =1
+ bk
)
∞
22) If
∑ (a
∞
∑a
)
+ bk is convergent, then so are
k
k =1
k
∞
and
k =1
FALSE: Consider ak = 1 and bk =
1
−1
k2
∞
∑(
k =1
∑b
k
.
k =1
)
∞
ak + bk = ∑
k=
1
converges but
k2
∞
∑b
k =1
k
does not.
∞
23) If lim ak = 0 , then
k→∞
∑a
is convergent.
k
n=1
1
= 0 but
k→∞ k
FALSE: lim
∞
∑
n=1
FALSE: This because
is divergent.
k =1
∞
24) One can prove that
1
∑k
1
converges by applying the Comparison Test using the series
n+8
1
1
< . If the bigger series diverges, which
n+8 n
∞
∑ n1 .
n=1
∞
∑ n1 does, then the smaller
n=1
∞
series can converge or diverge. To apply the comparison test for
∑ n +1 8 , note that
n=1
1
1
1
>
=
and
n + 8 n + n 2n
∞
∑ 2n1 diverges (constant multiple if the harmonic series)
n=1
∞
25)
∑n 1
n=1
cos x
is a convergent series for all x.
∞
FALSE:
∑n
n=1
1
cos x
∞
is a p-series. 0 ≤ cos x ≤ 1 for all real numbers x ⇒
∑n 1
n=1
cos x
is always a divergent
∞
p-series. On the other hand, unless x = kπ , where k is an integer, then the series
∑n 1
n=1
converges, provided that sec x is defined.
sec x
\